Which of these equations relates the pOH of a buffer to the p𝐾𝑏 of its weak base, analogous to the Henderson–Hasselbalch equation for weak acids?
a. p𝐾𝑏=pOH+p𝐾𝑏=pOH+log[acid]/[base]
b. p𝐾𝑏=pOH−log[acid]/[base]
c. p𝐾𝑏=pOH−log[base]/[acid]
d. p𝐾𝑏=pOH+log[base]/[[acid]

Question

Which of these equations relates the pOH of a buffer to the p𝐾𝑏 of its weak base, analogous to the Henderson–Hasselbalch equation for weak acids?

a. p𝐾𝑏=pOH+p𝐾𝑏=pOH+log[acid]/[base]

b. p𝐾𝑏=pOH−log[acid]/[base]

c. p𝐾𝑏=pOH−log[base]/[acid]

d. p𝐾𝑏=pOH+log[base]/[[acid]

Accepted Answer

well everyone. So in this video we're trying to find the correct relationship or mathematical relationship that relates R. P. O. H. Of the buffer to the P. K. B. Of the base components. So let's first go ahead and write out the equation or the equilibrium equation for this. So we're gonna go ahead and just put our base as just the letter B. So we have B. Again that's the base being acquis reacting with water which is in a liquid state to go ahead and give us B. H. Plus pro native base And A. O. H. So we get that R. K. B. Value is of course the concentration of our products over the concentration of the reactant. So on our product side we have then our B. H. Plus and R. O. H minus over. Let's see here our reactant we have to but we only enter in any Aquarius um reagents in our case that's just going to be our base. So be so we're gonna go ahead and manipulate this equation and just isolate for O. H minus If we do that we get that the concentration of age will be equal to KB multiplied with the concentration of our base over the concentration of B. H. Plus so now we can go ahead and take the log of both sides. Actually continue to write this in different color. So we're taking the log of both sides if we do so I apologize the negative log of both sides if we do so we get that the negative log of the concentration of R. O. H. Is equal to. Let's see the negative log of this big equation. So KB multiplied by the concentration of our base over. The concentration of B. H. Plus. We can use the properties of log now to go ahead and simplify this. So we get the negative log of our base minus the negative log of our concentration of B. H plus One more thing I actually forgot here let's move this over. We have of course our KB. two. Alright now again expanding the right side of our equation by using the properties of log we get negative log the concentration of H minus equaling two negative log of R. K. B minus the log of our base. And of course we're just adding the negative and negative to give us positive or plus log the concentration of R. B. H plus. So now we're rearranging this right side at least. So I'm gonna put are negative log of KB first then we're adding the log of the concentration of B. H plus minus the log of our base. So we see here two things we know that the equation to find our P. O. H. Is equal to the negative log of the concentration of O. H minus. And the equation to finding our Kb, R P K B is the negative log of well the concentration of our key. And so then go ahead and use those two and green to go ahead and simplify our equations here. So we have then our P O. H equaling two R P k B. And of course using the again properties log and simplified the two remaining logs. So we have the addition of our log of the concentration of R b H plus over the concentration of our base. So we take a look at the four possible options that we have here, the one that best or does match this. Let's see here, we're looking for our KB and adding the concentration of log B H plus and B. So we can see that our correct answer will be C. Which is that the p O H is equal to the PKb, adding the log of our fraction on numerator. We have the concentration of B H plus denominator, we have the concentration of our base and yeah, C is going to be my final answer for this problem.

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Key Concepts

Henderson-Hasselbalch Equation

pOH and pKb Relationship

Buffer Solutions